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A Genesis Project Based On A Six-fold Theory of Reality

Ian Beardsley

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Table of Contents

Abstract……………………………………………………………….3

Forward……………………………………………………………….4

1.0 Bridging the Microcosmos to the Macrocosmos…………5

2.0 Inertia As Proton-Seconds…………………………………….8

3.0 The Proton Radius………………………………………………13

4.0 Predicting The Life Elements………………………………….16

5.0 The Age of the Universe………………………………………..20

6.0 Summary…………………………………………………………..23

7.0 Overview…………………………………………………………..24

8.0 The Second Is The Metric………………………………………30

9.0 Kilogram-Seconds……………………………………………….35

10.0 Apophis……………………………………………………………39

11.0 Solving The System……………………………………………..42

12.0 The Proton Charge………………………………………………46

13.0 The Solar Magnetic Field……………………………………….48

14.0 Eightfold Symmetry……………………………………………..60

Appendix 1 Kinetic Energies Moon and Earth…………………….64

Appendix 2 Van Der Waals Radius…………………………………..65

Appendix 3 Chandrasekhar Limit……………………………………68!

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Abstract

We suggest the idea of a Genesis Project based on a six-fold theory of reality. Such a project

would be to solve the mystery of life-bearing worlds and systems so as to maintain them or,

even to transform dead worlds into living systems."

We develop the concept of intermediary mass somewhere between that of a proton and the

lower limit for a white dwarf star. From that we develop a constant k that bridges the

microcosmos with the macrocosmos. With this constant we predict the radius of a proton then

look at how it explains elements as based around 6-fold symmetry in the core life elements

carbon and hydrogen as a consequence of a theory for space, matter and time."

We ﬁnd that though the Ancients developed the second by dividing the the Earth rotation and

orbit, and the Moon’s orbit into a convenient calendar, they unintentionally created the duration

of second we have today in a way that conveniently describe the atom and the Earth-Moon-

Sun System. It conveniently takes the unit of one in terms them making computing easier in

our model. As such we suggest the second is a Natural Metric."

Our six-fold theory of reality for a Genesis Project predicts the life span of the Universe, the

length of the Earth day, the length of the Earth year, and explains the ratio of the Moon’s kinetic

energy to that of the Earth in terms of the ratio of the radius of a proton to its mass. At the end

we suggest the magnetic ﬁeld of the Sun may predict carbon, the core element of life

suggesting, it might play a role in a Genesis Project."

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Foreward

One of things I wanted to go into in this paper but didn’t is that you can speak of the structure of

the solar system even though it changes with time. This is important to understand when I refer

to sizes the Moon and the planets, and their orbital distances.

The whole object of developing a theory for the way planetary systems form is that they meet the

following criterion: They predict the Titius-Bode rule for the distribution of the planets; the

distribution gives the planetary orbital periods from Newton’s Universal Law of Gravitation. The

distribution of the planets is chiefly predicted by three factors: The inward forces of gravity from

the parent star, the outward pressure gradient from the stellar production of radiation, and the

outward inertial forces as a cloud collapses into a flat disc around the central star. These forces

separate the flat disc into rings, agglomerations of material, each ring from which a different

planet forms at its central distance from the star (it has a thickness). In a theory of planetary

formation from a primordial disc, it should predict the Titius-Bode rule for the distribution of

planets. When I speak of the state of the solar system I speak of this point toward which the

solar system has formed and not the small changes that happen over millions of years due to

mutual interference between the bodies. In fact, mutual interference has torn apart possible

forming planets resulting in the current distribution we have today, because the current

distribution is more or less stable. The asteroid belt is a good example of this — it is a location

where a planet cannot form due to harmonic (repetitive) action on the orbital period at its

distance by orbital periods of planets beyond it. In short we take the state of the solar system as

an inflection point between what it became, and what it might minutely be going away from in

billions of years after it dies and can no longer support life.

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1.0 Bridging the Microcosmos to the Macrocosmos

To bridge the microcosmos to the macrocosmos we introduce a constant k."

Warren Giordano wrote in his paper The Fine Structure Constant And The Gravitational

Constant: Keys To The Substance Of The Fabric Of Space, March 21, 2019:

In 1980, the author had compiled a series of notes analyzing Einstein’s geometric to kinematic

equations, along with an observation that multiplying Planck’s constant ‘ ’ by ‘ ’, where

‘ ’ is the Fine Structure Constant, and multiplying by yielded Newton’s gravitational

constant numerically, but neglecting any units.

This follows from what Warren Giordano noticed that

Equation 1.0

We can eliminate and introduce the 6 of six-fold symmetry by introducing Avogadro’s

Number

We make an equation of state for the periodic table:

We can say for any element

Where is the number of protons in the element, so for carbon

Thus by equation 1.0 we have

Equation 1.1.

1 + α

h(1 + α) ⋅ 10

= G

= 6E 23

ℍ = 1

gra m

atom

⋅ 𝔼 = 6E 23

𝔼

Z ⋅ 6E 23pr oton s

Z ⋅ gra m s

𝔼 =

Z ⋅ gra m s

Z ⋅ pr oton s

ℂ

ℂ =

6gra m s

6proton s

6(6E 23proton s)

6gra m s

(1 + α)

⋅ N

𝔼 = 6.0003kg

⋅

of 6 71

In the next step to bridge the microcosmos to the macrocosmos we create an intermediary

mass midway between the proton mass and the lower limit of mass for a white dwarf

star to form without collapsing into a blackhole star. It balances with its gravity by radiation

pressure alone, the so-called Chandrasekhar limit (See Appendix 3 for how it is derived):"

Equation 1.2

We make the approximation and define with the geometric mean:

Equation 1.3.

Giving

Equation 1.4.

Thus by equation 1.1 and equation 1.4 and using , we have

Equation 1.5.

We had

We can hone this by reintroducing 0.77 for 3/4

Thus precisely (Values of constants on page 8):

We have

We have honing our

M ≤ 0.77

ℏ

= 1.41 ⊙

0.77 ≈ 3/4

= Mm

ℏ

1/2

ℏ = h /2π

k =

8π

(1 + α) ⋅ N

𝔼

8π

1/2

= 67.9943kg

= 0.8660

0.77 = 0.8775

0.8775

0.8660

= 1.01328

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Thus since we said with our estimate

We have a honed value of

This is our constant for bridging the microcosmos with the macrocosmos. We define the Earth

to be the ground state for the solar system yielding our six-fold basis for Nature

Equation 1.6.

Where is the average orbital velocity of the Earth.

We note the ground-state for an electron in a hydrogen atom gives

Where is the potential energy of the electron and is its rest energy.

= (67.9943)(1.01328) = 68.897kg ≈ 69kg ≈ 70kg

k =

(1 + α)

⋅ N

𝔼 =

773.5

⋅

k =

(68.897kg)

(6.62607E − 34) ⋅

(1.007299)

6.67408E − 11

⋅ 6.02E 23 = 0.001268291s /m

= 788.4626

k v

= 6

k v

29790m /s

788.46m /s

= 6.145748

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2.0 Inertia As Proton-Seconds

We describe reality as being based in the six-fold. I really think Nature employs 6-fold

symmetry because I think it is the most dynamic being the product of the two smallest prime

numbers 2 and 3. Those are the smallest factors down to which anything can be reduced, so that

is why I think 6-fold is the basis of Nature, that the basis has to be reduced to the smallest

factors. As such we have 2x3=6, 3x3=9, 2x9=18 and 3x6=18. The periodic table of elements is

periodic over 18 groups, which means when you count to 18 the elements start over making

groups where they have similar properties falling into the same groups. These groups are

determined by electron configurations, for instance carbon is in group 14 which means since it

wants to have noble gas electron configuration, it gains 4 electrons to be the same as the noble

gas in group 18 of the same period. 18-14=4 meaning it ionizes as C4- so it has 4 electrons to

combine with 4 hydrogen atoms that are each H+ or so we can have two H atoms that can

combine with one oxygen atom which is O2- because oxygen is in group 16 meaning 18-16=2,

and so forth making life possible. We now need to show that the fundamental particles that

build reality are based on sixfold symmetry for their mass, size, and charge and we want to do it

in terms of gravity on the macroscale, thus it has to use G the universal constant of gravitation

and Planck’s constant h, that quantizes energy on the microscale. I find I can do this as such:

The natural constants measure the properties of space and time. We can list some of them here:

(Proton Mass)

(Planck Constant)

(Proton Radius)

(Gravitational Constant)

(light speed)

(Fine Structure Constant)

Gravity is a property of space measured by the universal constant of gravity, G:

Equation 2.1

Matter, or inertia, which measures matter’s ability to resist a force is for each particle (protons

and neutrons) we will suggest given by:

Equation 2.2

Which describes mass per meter over time, which is:

Equation 2.3

: 1.67262 × 10

−27

h : 6.62607 × 10

−34

J ⋅ s

: 0.833 × 10

−15

G: 6.67408 × 10

−11

c : 299,792, 459m /s

α :1/137

= q

= 1.6022E − 19coulom bs

= 8.988E 9

G = 6.674E − 11N

= 6.674E − 11

⋅ kg

kg ⋅ s

1.82E − 16

kg ⋅ s

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It must be adjusted by the fine structure constant . It is my guess the factor should be which

is 18,769.:

Equation 2.4

Because the fine structure constant squared is the ratio of the potential energy of an electron in

the first circular orbit to the energy given by the mass of an electron in the Bohr model times the

speed of light squared, that is it represents the ground state. It is

Equation 2.5

Here we are suggesting that the proton and neutron are the 3-dimensional cross-sections of a

hypersphere. Thus we consider the surface area of a proton, :

Equation 2.6

We take the square root to get meters:

Equation 2.7

We multiply that with the value we have in equation 2.4:

Equation 2.8

We find that the mass of a proton realizes if we divide this by six seconds:

Equation 2.9

That is hydrogen. We see that the element carbon manifests if we divide instead by 1 second:

Equation 2.10

Carbon (C) is the core element of life chemistry and it combines with hydrogen (H) to make the

skeletons of organic matter, the so-called hydrocarbons:

Equation 2.11

Equation 2.12

(1.82E − 16kg ⋅ s /m)(18,769) = 3.416E − 12kg ⋅ s /m

= 4π r

= 8.72E − 30m

= 2.953E − 15m

(2.953E − 15m)(3.416E − 12

kg ⋅ s

) = 1.009E − 26kg ⋅ s

1E − 26

6secon d s

= 1.67E − 27kg

1E − 26kg ⋅ s

1secon d

= 1E − 26kg = 6proton s = 6m

1E − 26kg ⋅ s

6secon d s

1E − 26kg ⋅ s

1secon d

of 10 71

It seems the duration of a second is natural. If it is, since it was formed by a calendar based on

reconciling the periods of the moon and the sun in the earth sky, it should be in the Earth-moon

orbital mechanics. I find it is, that (See Appendix 1):

Equation 2.13

That is, the earth day (86,400 seconds) times the kinetic energy of the moon to the kinetic

energy of the earth is about 1 second (about 1.2 seconds). The earth day changes very little, by

very small amounts over millions of years. The solar system has evolved towards this since the

explosion of life called the Cambrian, and will slowly decay away from it. But we need to derive

the second in terms of something else. For now we have the mass of a proton as:

Equation 2.14

This way of looking at things is to say matter is that which has inertia. This means it resists

change in position with a force applied to it. The more of it, the more it resists a force. We

understand this from experience, but what is matter that it has inertia? In this analogy we are

suggesting a proton is a three dimensional bubble embedded in a two dimensional plane. As

such there has to be a normal vector holding the higher dimensional sphere in a lower

dimensional space. (See Fig. 1) Thus if we apply a force to to the cross-section of the sphere in

the plane there should be a force countering it proportional to the normal holding it in a lower

dimensional universe. It is actually a 4-dimensional hypersphere whose cross-section is a

sphere. This counter force would be experienced as inertia. We are defining inertia as proton-

seconds, the action of subatomic particles over time. Our reasoning above in one equation is:

K E

moon

K E

earth

(Ear th Da y) ≈ 1secon d

18α

4πh

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Equation 2.15

Saying 1 second gives 6 protons is carbon (C)

Equation 2.16

And six seconds gives 1 proton is hydrogen:

From which instead of saying the left sides of these equations are seconds, we say they are

proton-seconds by not letting cancel with the bodies of these equations on the left, but

rather divide into them, which are in units of mass, giving us a number of protons. I say this is

the biological because as we shall see our equations are based on one second is 6 protons is

carbon, and 6 seconds is one proton is hydrogen, these making the hydrocarbons which are

the skeletons of biological life. We see this is a mystery of six-fold symmetry based around

biological life in the following computer program I wrote and its output:"

Is proton-seconds. Divide by time we have a number of protons because it is a mass divided by

the mass of a proton. But these masses can be considered to cancel and leave pure number.

We make a program that looks for close to whole number solutions so we can create a table of

values for problem solving.

By what value would you like to increment?: 0.25

How many values would you like to calculate for t in equation 1 (no more than 100?): 100

24.1199 protons 0.250000 seconds 0.119904 decpart

12.0600 protons 0.500000 seconds 0.059952 decpart

8.0400 protons 0.750000 seconds 0.039968 decpart

6.0300 protons 1.000000 seconds 0.029976 decpart

4.0200 protons 1.500000 seconds 0.019984 decpart

3.0150 protons 2.000000 seconds 0.014988 decpart

2.1927 protons 2.750000 seconds 0.192718 decpart

2.0100 protons 3.000000 seconds 0.009992 decpart

1.2060 protons 5.000000 seconds 0.205995 decpart

1.1486 protons 5.250000 seconds 0.148567 decpart

1.0964 protons 5.500000 seconds 0.096359 decpart

1.0487 protons 5.750000 seconds 0.048691 decpart

1.0050 protons 6.000000 seconds 0.004996 decpart

0.2487 protons 24.250000 seconds 0.248659 decpart

0.2461 protons 24.500000 seconds 0.246121 decpart

h 4πr

= 6proton ⋅ secon ds = carbon(C )

h 4πr

= 1proton ⋅ 6secon d s = h ydrogen(H )

h4πr

of 12 71

A very interesting thing here is looking at the values generated by the program, the smallest

integer value 1 second produces 6 protons (carbon) and the largest integer value 6 seconds

produces one proton (hydrogen). Beyond six seconds you have fractional protons, and the rest

of the elements heavier than carbon are formed by fractional seconds. These are the

hydrocarbons the backbones of biological chemistry. Here is the code for the program:

#include <stdio.h>

#include <math.h>

int main(int argc, const char * argv[]) {

int n;

ﬂoat value=0, increment,t=0, p=1.67262E-27, h=6.62607E-34,G=6.67408E-11,

c=299792459,protons[100],r=0.833E-15;

{

printf("By what value would you like to increment?: ");

scanf("%f", &increment);

printf("How many values would you like to calculate for t in equation 1 (no more than 100?): ");

scanf("%i", &n);

}

while (n>=101);

{

for (int i=0; i<n;i++)

{

protons[i]=((137*137)/(t*p))*sqrt(h*4*(3.14159)*(r*r)/(G*c));

int intpart=(int)protons[i];

ﬂoat decpart=protons[i]-intpart;

t=t+increment;

if (decpart<0.25)

{ printf("%.4f protons %f seconds %f decpart \n", protons[i], t-increment, decpart);

}}}}

We ﬁnd exactly our equation predicts the second as"

Equation 2.16. "

We said"

But that was using the average orbital velocities. We ﬁnd using aphelions and perihelions (See

Appendix 1)"

6α

h4πr

= 1.004996352secon ds

moon

Earth

(Ear th Day) = 1.2second s

K E

moon

K E

earth

(Ear th Da y) = 1.08secon ds

of 13 71

3.0 The Proton Radius

Thus we have the radius of a proton is given by carbon by evaluating at one second:

But to get that we have to multiply by one second and we need one second in terms of the atom

for a theory of the proton. I find we can do that…

Substitute for to get

Where is the Van Der Waals radius for a hydrogen atom (See Appendix 2). We have

now introduced the radius of a hydrogen atom . Our formulation of inertia as

proton seconds is a form of impulse. To change that to momentum we have to divide by a

second. This radius of the hydrogen atom is the Van Der Waals radius, which is the closest

distance between two hydrogen atoms noncovalently bound. It is 120 pm. Divide that by

where 1/k is our constant

And we find

We have our equation for the radius of a proton

We only need to multiply it by to have the right units, and we get!

4πh

= 8.288587 × 10

−16

m = 0.829f m

h 4πr

h 4π

t =

2α

h 4π

= 1.2E − 10m

⋅

8π

⋅

1 + α

⋅

𝔼

≈

⋅

8π

⋅

𝔼

3 2

⋅

6.626E − 34

6.674E − 11

⋅

18769

(1.6726E − 27)

⋅

1.2E − 10

6.02E 23

⋅

= 1.12secon d s

3 2

⋅

Gπ

⋅

𝔼

= 1.12secon d s

4πh

= 8.288587 × 10

−16

m = 0.829f m

t /ck = 1secon d

of 14 71

Equation 3.1.

Then suggest we picked up 9/8 in approximations which is close to one anyway so we write

Equation 3.2.

We form constants:

And we have the Equation:

Equation 3.3

We can say that Avogadro’s number is not an arbitrary number because it is such that there are

twelve grams of carbon and carbon is 6 neutrons plus 6 protons equals 12. We now want to add

to this section the equation of the mass of a proton.

Equation 3.4.

If our equation is right and we put it into natural units then the product should be close to

one:

Let us start with the units with which we are working:

2 ⋅

4π

⋅

𝔼

2π

⋅

𝔼

= 8.26935E − 16m ≈ 0.827f m

(

)

(

)

(

)

ℍ

(

)

k =

2π

= 6.93E − 9kg

𝔼

= 1.99E − 34m

= k

𝔼

2π

⋅

𝔼

G =

kg ⋅ s

h = kg ⋅

c = m /s

of 15 71

And convert these to proton-masses and proton-radii:

Now we find k in these units:

Thus we have:

Equation 3.5.

We will say our theory is holistic and that we are describing the proton radius in terms of the

whole of which it is a part, namely, the radius of a hydrogen atom, more specifically the Van Der

Waals radius, which is determined by hydrogen gas, or (See Appendix 3 for the theory of the

Van Der Waals radius)

G = 6.67408E − 11

kg ⋅ s

⋅ 1.67262E − 27kg(0.833E − 15m)

⋅ s = 193,131, 756

h = 6.62607E − 34kg ⋅

⋅

(0.833E − 15)

(1.67262E − 27kg)

= 5.71E 23

c =

(299,792, 459m /s)(1sec)

(0.833E − 15m)

= 3.6E 23

1.2E − 10m

0.833E − 15m

= 144,058

k =

2π

= 6.93E − 9kg

k =

(5.71E 23)(3.6E 23)

2π

(193131756)

= 4E18proton − m a sses

= k

𝔼

(4E18)(144058)

(6E 23)

5.76E 23

(6E 23)

= 0.96 ≈ 1

of 16 71

4.0 Predicting The Life Elements

We have that"

Equation 4.1 "

Equation 4.2. "

Which give the core life elements carbon and hydrogen, which are also the hydrocarbons that

make-up the skeletons of biological life chemistry:"

Equation 4.3. "

Saying one second gives 6 protons is carbon (C) and, six seconds gives 1 proton is hydrogen

(H):"

Equation 4.4. "

We have the six-fold symmetry in the constant k and the orbital velocity of Earth:"

Equation 4.5. "

We substitute Equation 4.2 and 4.5 into equation 4.3 giving:"

Equation 4.6"

Or we can let cancel with the numerator and not write in protons:"

Equation 4.7"

We see the radius of a proton to the mass of a proton is proportional to the kinetic energy of

the moon to the kinetic energy of the Earth, times the Earth rotation period (EarthDay) times the

square root of the Earth’s orbital velocity and equal by the constant square root k.!

6α

⋅

h4π

= 1secon d

Earth

moon

(Ear th Day) = (1secon d )

h4πr

= 6proton ⋅ secon ds = carbon(C )

h4πr

= 1proton ⋅ 6secon ds = hydrogen(H )

k v

= 6

⋅

h4π

⋅

earth

moon

⋅

Ear th Day

= k v

⋅ protons

⋅

h4π

= k

1/2

⋅

moon

earth

(Ear th Day) v

of 17 71

But the orbital velocity of the Earth is given by the Sun’s mass and Earth’s distance from the

Sun:"

Equation 4.8. "

But the Moon perfectly eclipses the Sun which means as seen from the earth the Moon

appears to be the same size as the Sun. This is because"

Equation 4.9. "

Where is the Earth orbital distance, is the Moon’s orbital distance, is the Sun’s radius,

and is the Moon’s radius.. Thus in terms of the moon and the Sun"

Equation 4.10. "

So we have from 4.8"

Equation 4.11. "

This means from 4.7"

Equation 4.12"

Interesting I found that the radius of the Sun to the Moon’s orbital radius in the ratio of a mass

of a gold atom to the mass of a silver atom which is about 1.8 or 9/5 by comparing the molar

mass of gold to that of silver. Interesting because gold and silver have been used since ancient

times as metals for making ceremonial jewelry and the Sun is gold in color and the Moon is

silver in color. We can write this"

Equation 4.13. "

Au (gold) and Ag (silver) are also some of the ﬁnest metals for electronic circuitry as they are

such ﬁne conductors. The above equation 4.12 has the radius of a proton and its mass

described by the kinetic energy of the Moon to that of the Earth perhaps because the Moon

makes life possible holding the Earth at its inclination to its orbit around the Sun allowing for

the seasons, and this ratio might be optimal for that. Naturally to optimize the success of life on

Earth this might have to be a factor of the Earth’s rotation which is given above by EarthDay.

⊙

≈

⊙

= G

⊙

⋅

h4π

= k

1/2

⋅

moon

earth

(Ear th Day) G

⊙

⋅

1/2

⊙

= 1.8

of 18 71

The orbital velocity of the Earth in the equation is determined by the Earth’s distance from the

Sun and the radius of a proton to the mass of a proton seems to determine that because

perhaps the velocity of the Earth determines its distance from the Sun and that in turn happens

to be the habitable zone, because it is the distance from the Sun where water can exist on

Earth in its liquid phase. Essentially equation 4.12 says how the kinetic energy of the Moon is in

the ratio of the radius of a proton to its mass, and it is that kinetic energy of the Moon that

makes like possible on Earth. Further the moon is composed of those same protons, and the

neutrons which have a similar ratio in their radii and mass. I would think this equation would be

central to any kind of “Genesis” project where we try to understand the secrets of making life-

bearing worlds so we can maintain or convert dead worlds like Mars to living worlds so we can

colonize them."

The length of the Earth day is related to the annual average temperature of the Earth, and as it

would turn out Mars, the one planet in our solar system we can colonize most eﬀectively, has a

day of the same length as that of the Earth, 24 hours."

Interestingly we can speak of the day of Mars and the day of the Earth interchangeably

because they are about the same, about 24 hours long for their rotations. We have to keep in

mind that Mars is smaller than Earth regardless that it completes a rotation in the same time.

The radius of Mars is 3,389,500 meters and that of the Earth is 6,378,100 meters. That is at its

surface Mars rotates through 246.5 m/s and at the surface of the Earth 463.8 m/s."

But interestingly, this ratio in size of Earth to the size of Mars is"

Equation 4.14 "

And the ratio of the Mars year to the Earth year is the same"

Equation 4.15 "

Interesting because the size of a planet has nothing to do with its orbital period. Orbital period

is a function of the distance of the planet from the star, not of its size. You could have an

asteroid at Mars orbit much smaller than Mars. The mass of Mars, of course is much lower too.

The mass of Mars is 6.39E23 kg and that of Earth is 5.97219E24 kg giving:"

Equation 4.16 "

The Earth is then about 9 times more massive than Mars. Equations 4.14 and 4.15 are to say"

Equation 4.17 "

Where is the Earth radius and is the Mars radius, and is the Earth year and is the

Mars year. This is interesting because it is similar to the parameters of the moon with respect

to the Earth and Sun that we suggest is a mystery that is a message from unknown forces. We

had"

6,378,100

3,389,500

= 1.88

687

365.25

= 1.88

5.97219E 24

6.39E 23kg

= 9.34615

earth

mars

earth

of 19 71

Equation 4.18 "

Where is the earth orbital radius, is the Moon’s orbital radius, is the Sun’s

radius, and is the radius of the Moon. Indeed the mystery of the Moon could speak to us

of the mystery of our existence even to our ancient ancestors ﬂaking spearpoints from stone

around the ﬁre a million years ago because they could witness the Moon perfectly eclipsing the

Sun, as much as it could be a message to us today. But in 15.4 where the Earth size to the

Mars size equals the Mars year (journey across the sky) to the Earth Year (journey of the Sun

across the sky) would require us to know the size of the Earth and the size of Mars, so if it is a

message like the lunar message, it would only be readable later in the human story."

Given Kepler’s Law of Planetary motion"

Then we have"

Equation 4.19. "

Bear in mind, interestingly, the ratio in size of Earth to the size of Mars is"

And the ratio of the Mars year to the Earth year is the same"

But as well and we will call it equation 4.20"

Equation 4.20. "

Where Au and Ag are the masses of gold and silver atoms. The 1.8 is close to 1.88."

earth

moon

⊙

moon

earth

moon

⊙

moon

4π

= r

(

mars

earth

)

4π

(

moon

⊙

)

6,378,100

3,389,500

= 1.88

687

365.25

= 1.88

⊙

moon

= 1.8

of 20 71

5.0 The Age of the Universe

We have the possible core equation for a so-called Project Genesis giving the ratio of the radius

of a proton to its mass the kinetic Energy of the moon to that of the Earth:

We now see the same equations predict the age duration of the Universe in the standard model.

That is we start with the following two equations:

Eq 5.1.

Eq. 5.2

We want to factor 5.1 into a ratio between kinetic energies times a time as we have in 2.2. We

write 2.1 as

And noticing

Which gives

And we know

And have

⋅

h 4π

= ⋅

K E

moon

K E

earth

(Ear th Da y) k v

6α

⋅

h 4π

= 1secon d

K E

moon

K E

earth

(Ear th Da y) = 1secon d

6α

⋅ Gc

4πh

1secon d

= K E

36α

⋅

⋅ KE

4π

1secon d

K E =

4π

36α

⋅

6α

h 4πr

of 21 71

We also notice that obviously by dimensional analysis

We have the kinetic energy and we need another kinetic energy to form the ratio:

We have the intermediary mass and the constant k inverted. We can write

This is

=3.03E56 seconds

But we also formulated

4π

36α

⋅

(0.833E − 15)

⋅ (1.67262E − 27) = 1.413E − 49J

4π

36α

⋅

= K E

⋅

(K E

moon

)

= 4.6654

K E = m

⋅

K E

(t im e) = aT im e

K E = m

(

)

K E = (68.897kg)(788.4626m /s)

= 42,831,358Joules

K E

(t im e) = (aT im e)

(

)

(1secon d ) =

(42,831,358J )

(1.4130E − 49J )

(1secon d )

of 22 71

So we can write

1 Earth year = (365.25)(24)(60)(60)=31557600 seconds

The universe is theorized by standard models to die in 100 trillion years, which is when the last

stars born will die out. This is exactly 1E14 years

We see the moon is connected to how old the universe is theorized to become.

6.62607E − 34

6.67408E − 11

⋅

299792459

1.67262E − 27

= 1.599298E 29J

⋅

= 1.599298E 29J

⋅

(

)

(1secon d ) =

1.599298E 29J

42831358J

= 3.734E 21secon d s

3.734E 21s

31557600s

= 1.1832332E14years ≈ 1E14years

of 23 71

6.0 Summary

We would like to present some things clearly:

Thus

If is 1.00 seconds even. In terms of our theory

Or as another estimate:

Because

4πh

= 8.288587 × 10

−16

m = 0.829f m

6α

h 4πr

= 1.004996352secon d s

t =

2α

h 4π

⋅

8π

⋅

1 + α

⋅

𝔼

≈

⋅

8π

⋅

𝔼

3 2

⋅

6.626E − 34

6.674E − 11

⋅

18769

(1.6726E − 27)

⋅

1.2E − 10

6.02E 23

⋅

= 1.12secon d s

3 2

⋅

Gπ

⋅

𝔼

= 1.12secon d s

4πh

= 8.288587 × 10

−16

= 0.829f m

π16α

⋅

3 2

𝔼

⋅

= 1.12secon d s

πα

⋅

𝔼

⋅

= 1.067secon d s

3 2

= 0.265165

= 0.23570

0.23570

0.265165

= 0.888888

of 24 71

7.0 Overview

If the universe was born from a big explosion called “the big bang” then it is expanding and the

distances between galaxies is growing. This does not mean there is an edge to the universe —

space could continue on infinitely in all directions but there could be a horizon beyond which

there is no content.

There can be an observable edge to the universe, but it is more of a horizon, just as there is a

horizon to the ocean, we know beyond the horizon there is more water. Since the universe was

born 13.82 billion years ago in the Big Bang we can only see those galaxies whose light has taken

less than 13.82 billion years to reach us. These galaxies form a sphere around the Earth called

“the observable universe”. It is 92 billion light years in diameter because in its beginning the

universe inflated much faster than the speed of light.

The Schwartzchild radius:

Gives the event horizon of a black hole. I have heard it said if the mass, M, is the mass of the

universe, that radius is the edge of the observable universe that, the observable edge of the

universe is actually the event horizon of a black hole and that then the Universe is a giant black

hole and that we are inside of it. I would like to call this to question, but first let’s discuss the

types of possible universes.

The pivotal parameter describing the universe is 𝛀 (omega) which is the average matter density

of the universe divided by a critical value of that density. Whether omega is less than1, 1, or

greater than 1 determines whether the Universe is open, flat, or closed. If matter is mostly inert

as in the dust models, there is a particular fate for each omega. Since 1998 the observations in

supernovas indicate the universe is accelerating in its expansion. To explain this cosmologists

hypothesize the existence of dark energy which can be any field with negative entropy causing

negative pressure. If 𝛀<1 then the Universe is open and the fate of the universe is in its heat

death, the burning out of the last stars born. This is also true of a flat universe if it has the

hypothesized dark energy, because then it accelerates to escape collapse. The fate of the universe

depends on its density and today most evidence points to it will expand indefinitely. In this

scenario there is enough a supply of gas for stars to form for 10¹² to 10¹⁴ years. 1 to 100 trillion

years. That is the last stars born would die in 10¹⁴ years. There are models that suggest the

universe is eternal where as stars die more matter comes into existence from which more stars

are continuously forming.

Now let’s call into question that the edge of the observable universe is the Schwartzchild radius,

the event horizon of a blackhole.

I had seen a documentary where they said there was a cosmologist working on a theory that the

edge of the observable universe is the event horizon of a blackhole. I thought okay as weird as

that seems, and I say weird because that means either the Earth is at the center of this blackhole,

or if not, mysteriously wherever you are in the universe is the center. I accepted this as possible

because the more we learn the stranger the universe gets. Indeed a long time ago Niels Bohr said

that if you haven’t realized that quantum mechanics says everything that makes up the universe

can’t be considered real, you have missed its point. And indeed if you think about quantum

mechanics, this rings true, so yes I was able to accept this about the edge of the universe, as

2GM

of 25 71

possible. However, taking a second look we see what is going on here may be circular

reasoning…

Einstein’s equations give the relationship between spacetime and matter content of the

Universe. From this Friedmann gave us his equations that are at the heart of cosmology, which

are functions of time, the scale-factor R, the matter density of the universe , and the pressure p

due to radiation produced by the stars, or by the galaxies because they contain the stars:

Where

for a 3-sphere, for 3-space, and for a 3-pseudosphere. These Friedmann

equations give the critical density of the universe as

Where is the Hubble constant which is the expansion rate of the universe. Current estimates

are that it is . A Hubble sphere is given, which is given by

Since the density of the universe is very close to the critical density which is

, then since

So this is close to the observable edge of the Universe. But the radius of the Universe is much

larger than 13.8 billion light years, because it is expanding we say it is 46.508 billion light years.

However we should look at where we got this mass we used, it came from the Hubble constant

and we used . What we did is:

R′ ′

(

R′

)

+ 8π p = 0

(

R′

)

8πρ

R′ ≡

κ = + 1

κ = 0

κ = − 1

8π G

71k m /s /Mpc

299792549m /s

71,000m /s /Mpc

= 4222.429Mpc

pc = 3.26l y

Mpc = 3.26 E6ly

= 13.8E 9ly

≈ 8.6E − 30g /cm

= 8.6E − 27kg/m

= 1.38E10l y = 1.3E 26m

M = ρ

π R

= 8E52kg

2GM

= 2

(6.67E − 11)(8E52kg)

299792459

= 1.18742E 26m = 12.55E 9ly

of 26 71

Which is

Which is the critical mass given by the Friedmann equations. So, we have gone in a circle, I

think. Let us use our value of 46.508E9 ly = 4.4E26m instead of 1.3E36m:

That is larger by a factor of 36.255. Not to say we aren’t in a blackhole, it just can’t be at the edge

of the observable universe, the radius of the Hubble sphere. Now with this brief sketch of

cosmology that we have done, let us talk again about some of the conclusion in this paper. We

have the second may be a natural duration. We suggest this because we found"

Or…"

2GM

= 2

(

)

8π G

M = ρ

π R

= 3.0686E54kg

2GM

= 2

(6.67E − 11)(3.0686E54kg)

299792459

= 4.55E 27m = 4.81E11l y

6α

⋅

h 4π

= 1secon d

K E

moon

K E

earth

(Ear th Da y) = 1secon d

⋅

(

)

⋅

K E

moon

K E

earth

(Ear th Da y) = Li feSpa nUniverse

⋅

(

)

⋅

6α

⋅

h 4π

= Li feSpa nUniverse

π16α

⋅

3 2

𝔼

⋅

= 1.12secon ds

πα

⋅

𝔼

⋅

= 1.067secon ds

of 27 71

But let us ask why did the unit of a second come out as natural in the structure of Nature and

the Universe. We got it from ancient times by dividing the Earth day (1 rotation of the Earth) into

24 hours and that into 60 minutes and that into 60 seconds which is to divide it into"

The smallest primes, smallest factors down to which we can reduce a fraction are, 2 and 3. We

have"

, "

And there are 18 groups in the periodic table of elements because the properties of the

element are periodic over cycles of 18. We have"

, "

Is the golden ratio."

And degrees come from in circle giving"

Is unity."

(24hours)(60 minutes)(60second s) = 86400secon ds

1Rotation

86400s

1Day

86400s

2 ⋅ 3 = 6

3 ⋅ 6 = 18

86400

= 43200

86400

= 28800

86400

= 7200

86400

= 14400

7200

100

7200

= 72

2 cos(72

∘

) = Φ =

5 + 1

360

∘

86400

360

= 240

2 cos(240

∘

) = − 1

86400

= 14400

14400

= 1800

86400

= 10800

= 0.75

of 28 71

So we are dealing with the reconciliation of the six-fold with the eight-fold. This suggests to me

the periodic table of the elements, and I had found something in earlier work that I had done

this is group 14 in the periodic table of elements its ﬁrst two elements carbon, the core element

of life and silicon (Si) the core element of Earth (or sand) which is silicon dioxide ( ). Let’s

look at this on the next page…."

SiO

of 29 71

We see that the atomic radius of silicon the core element of artiﬁcial intelligence (transistor

technology) ﬁts together with the core element of biological life carbon if the silicon is taken as

inscribed in a regular dodecagon (12 sides) and the carbon is taken as inscribed in a regular

octagon (8 sides). We have:"

, , "

Apothem: "

For a regular dodecagon:"

The radius of a silicon atom is Si=0.118nm and that of carbon is

C=0.077nm:"

This has an accuracy %"

Twelve is an expression of six in that it allows you describe it as two three’s and three two’s.!

D = 1 + 2x

+ x

= 1

x = 2 /2

D = 1 + 2

a = (1 + 2)/2 = 1.2071

a =

s/2

tan(θ /2)

0.5

tan(15

∘

)

= 1.866

0.118

0.077

= 1.532

1.866

1.2071

= 1.54585

1.532

1.54585

100 = 99

of 30 71

8.0 The Second Is The Metric

We have "

8.1. "

8.2.

, and . In both 11.1 and 11.2 we have

kinetic energy over kinetic energy times one second equal to a pivotal time period in Nature, the

first the Life Span of the Universe, and the second the Earth Day, period of rotation of the Earth.

In equation 11.1 is given by the white dwarf star:

And

Is the Chandrasekhar limit for the mass of a star to not collapse into a blackhole so it can

become a white dwarf star. The Life Span of the Universe is perhaps related to the Schwartzchild

radius which gives the event horizon of a blackhole star. It is

8.3.

If is the mass of the Universe

8.4.

Where is the radius of the Universe and is the critical density of the Universe that

determines whether the universe is open, flat, or closed. It is

8.5.

Where

⋅

(

)

(1secon d ) = Li feSpanUniverse

K E

Earth

K E

moon

(1secon d ) = Ear th Da y

⋅

= Kin et icEn erg y

(

)

= Kin et icEn erg y

= Mm

M = 0.77

8π

2GM

M =

π R

8π G

of 31 71

8.6.

Is the radius of the Hubble sphere is observable edge of the Universe. is the Hubble constant

is the expansion rate of the Universe which is thought to be approximately 71km/s/Mpc. And,

the critical density is approximately the average density of the Universe. We can write 11.1

8.7.

Using where .

We have

EarthDay=(24)(60)(60)=86400 seconds

LifeSpanUniverse=1E14 Years = 3.734E21 seconds

KE Moon=3.428E28J

KE Earth=2.7396E33J

Kinetic energies used from aphelions and perihelions calculated earlier in this paper. We have

Using our honed value for k.

=(5.354E-34)(1.5657E-10)=8.3827578E-44

Multiplying these last two:

(3.96631kg)(8.3827578E-44)=3.32486E30 kg

This mass should equal that of a white dwarf star (Its lower limit not to collapse into a

blackhole). There can be some play as we are suggesting the moon is an indicator somehow of all

that is going on in the Universe, and it only has to be at approximately the right orbit to do this.

We have

Which makes the equation pretty accurate in this form considering we are comparing something

like the Earth day to something as immense as the LifeSpanUniverse. We are going to want to

look at:

= R

⋅

(

Ear th Da y

Li feSpa nUniverse

)

(

K E

moon

K E

earth

)

= M

WhiteDwar f

= Mm

M = M

WhiteDwar f

⋅

(

788.4626

)

(1.6726E − 27)

(6.626E − 34)

(6.674E − 11)

(299792459)

(2.58747E − 12)(2.1371E 80)(9.926581E − 47)(7.25979E50) = 3.96631k ilogra m s

(

Ear th Da y

Li feSpa nUniverse

)

(

K E

moon

K E

earth

)

(

86400

3.734E 21

)

(

3.428E 28J

2.7396E 33J

)

2.86416E 30kg

3.32486E 30kg

= 86 %

of 32 71

Where is the Chandrasekhar limit. We now write

Where is the Schwartzchild radius is the is the edge of the observable universe. If

we have

Then

Yields

Which is the critical density of the Universe. We write 8.8

8.8.

⋅

(

)

(1secon d ) = Li feSpa nUniverse

8π G

⋅

(

)

= R

8π G

⋅

(

)

= R

8π G

⋅ ρ

= 1

8π G

⋅

(

)

(1secon d ) = Li feSpa nUniverse

= 71,000m /s /Mpc

(3.26E6ly /Mpc)(9.461E15m /ly) = 3.1E 22m /Mpc

= (71000m /s)(3.1E 22m) = 2.3E − 18s

−1

3(2.3E − 18)

8(3.141)(6.6741E − 11)

= 9.461E − 27kg/m

of 33 71

Thus we have

Where is a white dwarf star. We have

We suggest for some mass , we have

It is given by

8.9

For all practical purposes this is the mass of the Moon, which is exactly 7.34767E22kg.

Conclusion next page…

⋅

(

)

(1secon d ) = Li feSpa nUniverse

K E

Earth

K E

moon

(1secon d ) = Ear th Da y

⋅

(

)

(1secon d ) = 1Ear thYear

M =

⋅

(1Ear thYear)

(

)

(1secon d )

⋅

6.62607E − 27

6.67408E − 11

⋅

(299792459)

1.67262E − 27

= 1.599E 36J ≈ 1.6E 36J

= (788.4626)

= 621,673.27

1Ear thYear = (365.25)(24)(60)(60) = 31557600s

M =

1.599E 36

621673

⋅

31557600

= 8.15E 22kg

8.15E 22kg

7/34767E 22kg

= 1.10919516 ≈ 1.12m oon s

of 34 71

Thus we have

Where is a white dwarf star and gives LifeSpanUniverse.

Where Earth-Moon-Sun orbital parameters give the Earth rotation period.

Where is the mass of the moon gives the Earth orbital period. We have already formulated

From the radius of a proton and the mass of proton , our theory panned out: The second is

a unit of time that expresses the cycles of the Universe, it is the metric (measures thing

conveniently). I think next we want to look at

⋅

(

)

(1secon d ) = Li feSpa nUniverse

K E

Earth

K E

moon

(1secon d ) = Ear th Da y

⋅

(

)

(1secon d ) = 1Ear thYear

6α

⋅

h 4π

= 1secon d

8π G

⋅

(

)

= R

8π G

of 35 71

9.0 Kilogram-Seconds

One of the things we have done is said since the Chandrasekhar limit can be taken as 1.44

solar masses, and the mass of the Sun is 1.98847E30kg, and the mass of a proton is

1.67262E-27kg, that our intermediary mass is"

We also computed"

Where came from the approximation of (3/4) for 0.77 in

Which we ultimately took as our value for the honed :

We have

We have honing our

We developed a concept of proton-seconds. Now we develop a concept of kilogram-seconds. To

do this we look at our equations for the cycles of the Universe:

9.1.

9.2.

= (1.44)(1.98847E 30kg)(1.67262E − 27) = 69.205kg

8π

1/2

= 67.9943kg

3/2

= Mm

= 0.77

ℏ

1/2

= 0.8660

0.77 = 0.8775

0.8775

0.8660

= 1.01328

= (67.9943)(1.01328) = 68.897kg ≈ 69kg ≈ 70kg

⋅

(

)

(1secon d ) = Li feSpa nUniverse

K E

Earth

K E

moon

(1secon d ) = Ear th Da y

of 36 71

9.3.

9.4.

We consider

9.5.

9.6.

And write

9.7.

Which is

9.8.

Where we remember

9.9.

Can be taken as

9.10.

If cancels with the numerator.

9.11.

9.12.

⋅

(

)

(1secon d ) = 1Ear thYear

6α

⋅

h 4π

= 1secon d

⋅

6.62607E − 27

6.67408E − 11

⋅

(299792459)

1.67262E − 27

= 1.599E 36J ≈ 1.6E 36J

= (788.4626)

= 621,673.27

⋅

(

)

(1secon d ) = 2.5721E 30kilogr am ⋅ secon ds

⋅

⋅ t

⋅ k

= 2.5721E 30kg ⋅ s

6α

⋅

h 4π

= proton ⋅ secon ds

6α

⋅

h 4π

= 1secon d

⋅

6α

⋅

⋅ k

h 4π

= 2.5721E 30kilogr am ⋅ secon ds

6α

⋅

h 4π

= proton ⋅ secon ds

of 37 71

We can now take 9.8 and divide by the mass of the moon:

(365.25)(24)(60)(60)=3157600 seconds is:

Meaning we have the Earth year to 90% accuracy. We can also divide by the mass of the Earth:

And we note as we said in the beginning of this paper the moon orbits the earth inclined 5

degrees to the Earth orbital plane (That is, to the ecliptic). In days the Earth goes about 5

degrees around the Sun because the earth goes through about 1 degree a day because there are

360 degrees in a circle and 365.25 days in a year. We can also divide kilogram-seconds by the

mass of the Sun to get:

And find it is very close to one second. We in fact have that if we use instead of aphelions and

perihelions to calculate the kinetic energies of the moon and earth, we can use their average

orbital velocities we have (Appendix 1):

Equation 9.13.

As opposed to

Equation 9.14.

Which allows us to write very accurately

⋅

⋅ t

⋅ k

= 2.5721E 30kg ⋅ s

2.5721E 30kg

moon

2.5721E 30kg ⋅ s

7.34767E 22kg

= 35005654.85s

31557600s

35005654.85

= 90 %

2.5721E 30kg

earth

2.5721E 30kg ⋅ s

5.972E 24kg

= 430693.2351s

430693.235s

31557600s

= 0.013647845years

73.27

c ycles

year

365.25d ays

73.27

= 4.98498d ays ≈ 5d ays

2.5721E 30kg

⊙

2.5721E 30kg ⋅ s

1.989E 30kg

= 1.29316s

K E

moon

K E

earth

(Ear th Da y) = 1.2secon ds

K E

moon

K E

earth

(Ear th Da y) = 1.08secon ds

of 38 71

Equation 9.15.

As we can see while kilogram-seconds deals with Natural cycles and periods

Proton-seconds are structured around the core element of life, carbon, and the most abundant

elements in life carbon and hydrogen and in the most abundant element in the Universe,

hydrogen, as founded on six-fold symmetry characterized in the hydrocarbons, the skeletons of

biological life and organic chemistry:

⋅

⊙

⋅ k

K E

moon

K E

earth

(Ear th Da y)

⋅

(

)

(1secon d ) = Li feSpa nUniverse

K E

Earth

K E

moon

(1secon d ) = Ear th Da y

⋅

(

)

(1secon d ) = 1Ear thYear

6α

⋅

h 4π

= 1secon d

h4πr

= 6proton ⋅ secon ds = carbon(C )

h4πr

= 1proton ⋅ 6secon ds = hydrogen(H )

of 39 71

10.0 Apophis

The reason that motivated me to write this paper was explained in the opening section, but not

in its more speciﬁc details. It was explained that the Moon seems to give us a hint that we are

here for a reason, like it perfectly eclipses the Sun, and that may have been a message to our

ancient ancestors, as well as to us today. Or in that the Moon holds the Earth’s inclination to

the ecliptic, that keeps the climate right to support life. These things were already known, but I

discovered something that made me pursue this as a paper, and that was in that both the

proton and the Moon deﬁne that duration of a second nearly perfectly, which I have found in

many ways make it a good unit for measuring Nature accurately. I call it a “Natural Metric”. In

light of how we got the second makes this an even deeper mystery; We got it from the way the

ancients divided up the rotation of the Earth and the Earth’s orbit to make a calendar that

works well. But I pursued this paper for an even deeper reason than mentioned: I was

wondering in light of negative forces in the world and positive forces in the world whether or

not we were slated for success as a species from the outset, and the Moon in light of what we

already knew, and what I found, gave me good reason to pursue the project. It panned out so

well that I began to feel there was something more than certainly going on here. Among other

things, like a theory for reality, I found what seems to be life cycles of the Earth and Universe. I

found"

⋅

⊙

⋅ k

K E

moon

K E

earth

(Ear th Da y)

⋅

(

)

(1secon d ) = Li feSpa nUniverse

K E

Earth

K E

moon

(1secon d ) = Ear th Da y

⋅

(

)

(1secon d ) = 1Ear thYear

6α

⋅

h 4π

= 1secon d

h4πr

= 6proton ⋅ secon ds = carbon(C )

h4πr

= 1proton ⋅ 6secon ds = hydrogen(H )

of 40 71

I decided to apply this to Apophis, an asteroid we thought might hit Earth in 2029 that could

have either wreaked havoc, or even brought about our extinction. Hence its name from the

Greek meaning “mayhem” and “chaos”. It is going to be in 2023 the nearest miss in history,

coming between us and the Moon and even within the orbits of our geosynchronous satellites. It

has an orbit of about year and comes near to Earth every seven years. We want to see if the

Moon predicts such an asteroid (of 7 years) with equation 12.3. This gives

The mass of the Moon is

But is predicted by equation 12.3 to be 8.15E22kg. Thus we divide this by our answer above for

This may be the 7 years, where the year is one orbit of the Earth and as well approximately that

of Apophis. Much more needs to be done, but we seem to be doing well with these equations of

Natural cycles so far. Of course with the second as a Natural Metric, the year is Natural as well

so the easiest way to do this is in equation 12.3 since one lunar mass gives 1 Earth year, so will

the 7 years of Apophis near miss give seven lunar masses. But considering the system of

equations given above, the question becomes why 7 years gives 7 lunar masses. We have to solve

this system of equation above, further. We only know Apophis passes between the Earth and the

Moon in 2023. In this paper I offer it up with more editing and correction of typos, and the

above suggestion of looking further into the system of equations for Apophis. This makes us

suggest mass is inversely proportional to time:

M =

⋅

(1Ear thYear)

(

)

(1secon d )

⋅

6.62607E − 27

6.67408E − 11

⋅

(299792459)

1.67262E − 27

= 1.599E 36J ≈ 1.6E 36J

= (788.4626)

= 621,673.27

1Ear thYear = (365.25)(24)(60)(60) = 31557600s

7Ear thYear = 220903200s

(1.599E 36J )

(220903200s)(621673.27)

= 1.16435E 22kg

= 7.34767E 22kg

Apophis =

8.15E 22

1.16435E 22kg

= 6.9996 ≈ 7

T =

of 41 71

Where T is in Earth years and M is in Lunar masses. This is a good clue for solving the framing

of our theory further that the Moon is the key to the Universe. And we are going to to that now

in the next section.

of 42 71

11.0 Solving The System We now undertake since the second is Natural, to show the Earth

year is as well because the Earth goes around the Sun not just in year, but in a second in

another sense. That is we can make a variable and say:"

Equation 11.1 "

We start with"

Equation 11.2

We write

Equation 11.3

Equation 11.4

Where is the mass of the Earth, and is the mass of the Moon. We now substitute for

the mass the mass of Jupiter because it significantly carries most of the mass of the solar

system excluding the Sun. It is 1.899E27 kg. We have

Equation 11.5.

This is about 1 Earth day because 1 Earth day=86400 seconds. We want to suggest that in our

scenario that the Natural cycles in time are inversely proportional to mass:

Equation 11.6

Because if , then . If the equation is:

Equation 11.7

Where M and T are in lunar masses and Earth years.

Let . We have the mass of Jupiter in Lunar masses is

= (1secon d,1secon d /1year, . . . )

⋅

(

)

(1secon d ) = 1Ear thYear

⋅

⋅ k

= 2.5721E 30kg ⋅ s

⋅

(

)

(1secon d ) = 1Ear thYear ⋅

= 2.1E 32kg ⋅ s

⋅

⋅ k

⋅

2.1E 32kg ⋅ s

1.899E 27kg

= 110584.5secon d s

∝ T

M = M

= 1year = T

⋅

= T

M = M

1.899E 27kg

7.34767E 22kg

= 2.5845E4Lu n arMasses

of 43 71

This is about one day since there are 365.25 days in a year. We already said

And we have

Is about one day because there are 365.25 days in a year. Thus we equate the latter two:

But, the expression on the left gives its answer in Earth years and the expression on the right

gives the answer in seconds. Let convert seconds to Earth years. Let

We have

So since the Earth orbital period and

We have

Again we have, but we divide by to put it in lunar masses on the left and let

Equation 11.8.

⋅

81.27

2.5845E4

= 0.00314years

⋅

⋅ k

⋅

2.1E 32kg ⋅ s

1.899E 27kg

= 1.2799d ays

⋅

81.27

2.5845E4

= 1.15d ays

⋅

= t

⋅

⋅ k

⋅

= (365.25)(24)(60)(60) = (1secon d )/31557600secon ds /year

⋅

⋅ k

⋅

= 110584.5secon d s

110584.5s

31557600s

= 0.00350years

31557600 = T

⋅

= 0.00314years

0.00350years ≈ 0.00314years

= 1secon d /T

⋅

1secon d

⋅

⋅ k

⋅

of 44 71

This is interesting because , is not a second but the Earth orbital period in

seconds. We have

Equation 11.9

This says

Equation 11.10

Let’s verify it, we wrote

We have

The measured mass of the moon is 7.34767E22kg, We are close by

accuracy

Thus we see

Equation 11.10 is our original equation, Equation 11.2

We can write it

But now

= 1secon d /T

1 =

⋅

⋅ k

⋅

1secon d

⋅

⋅ k

⋅

1secon d

⋅

⋅ k

= 2.5721E 30kg ⋅ s

2.5721E 30kg ⋅ s

31557600s

= 8.15E 22kg

7.34767E 22kg

8.15E 22kg

= 90 %

= 1secon d,1secon d /year, . . .

⋅

(

)

(1secon d ) = 1Ear thYear

⋅

(

)

⋅ t

= 1Ear thYear

= 1secon d,1secon d /1year, . . .

of 45 71

Because in lunar masses and Earth years it is to say

Since

Equation 11.11

Because

Equation 11.12

In the process we found out that

Where the mass of Jupiter gives one Earth day is one rotation of the Earth. We already had that

the mass of the Moon gives one orbit of the Earth:

We see the Solar system is like an atom, which is miraculously quantized in terms of the second,

which describes the atom and the Moon:

= T

1 =

⋅

⋅ k

⋅

1secon d

⋅

⋅ k

⋅

1second

= T

⋅

⋅ k

⋅

secon d

2.5721E 30kg

(7.34767E 22)(31557600s)

= 1.109 ≈ 1

⋅

⋅ k

⋅

= 1Ear th Da y

⋅

(

)

(1secon d ) = 1Ear thYear

6α

⋅

h 4π

= 1secon d

K E

Earth

K E

moon

(1secon d ) = Ear th Da y

of 46 71

12.0 The Proton Charge

I construct the electric field as such (Fig. 2): is one component of . Events in

are through time with components and where c is the speed of light. The

electrons and the protons and in the presence of one another cross into giving

themselves acceleration in the and directions.

We have

We suggest

Equation 12.1

The position of changes in the space of and has travelled to a place in

time, as well. Since ct=meters there is a g such that . Thus since

ℝ

(x, y, z)

ℝ

(x, y, z)

−x



0 ct

0 0 ct

= (ct

− ct

)



··

x ∝ (ct

− ct

)



ℝ

(x, y, z)

t (0,ct

, ct

)

··

x = g(ct

− ct

)



g ⋅ m =

Fig. 2

of 47 71

and that is . Thus g is frequency squared ( ) and we suggest it is

derived from the separation between the charges and . That there is some velocity v such

that . We call upon our equation for 1 second:

Letting v= from our equation for k

And we have

Equation 12.2

We get

Equation 12.3 !

g =

⋅

⋅ s =

f =

(

)

h 4π

= 1secon d

(

)

(c)

h 4π

⋅

(c)

h 4π

= k

⋅

k =

8π

(1 + α) ⋅ N

𝔼 =

773.5

= 773.5

m = m

(

⋅

)

h 4πr

= q

h 4πr

(6.626E − 34)(4π)(0.833E − 15)

(6.674E − 11)(299,792, 459)

= 5.37E − 31

q = (0.033)

36(18769

)

773.5

(5.37E − 31) = 9.1435E − 19C

= ⋅

9.1435E − 19C

1.602E − 19C

= 5.71protons ≈ 6protons

of 48 71

13.0 The Solar Magnetic Field

We model the formation of the solar system from a slowly rotating gas cloud, a nebula of

gaseous molecules, that collapses into a flat disc with a protostar at its center. The star turns on

and blows lighter elements far away, like hydrogen and helium, from which form the gas giants,

like Jupiter and Saturn, and the heavier elements stay closer in, like iron and silicates, from

which form the terrestrial planets like Venus, Earth, and Mars form. There are basically three

factors that determine its structure, the inward gravity, the pressure gradient outward which

balances with the inward gravity, and the outward inertial forces from the planets’ orbits. The

flattened rotating disc is broken up into rings each that has a mass spread out over it from which

the planets form. We estimate the ring associated with the Earth, had in its lower limit 230 earth

masses spread over it for the Earth to form. We further estimate that the Venus ring had a mass

spread over it of 230 Venus masses for Venus to form, and the Mars ring similarly had 230 Mars

masses spread out over it for Mars to form. The asteroid belt had about 200 of it masses, and the

Jovian planets 5, 8, 15, and 20 masses of each respectively. For Mercury it requires a factor of

about 350 because it is mostly iron condensations with incomplete silicon condensations.

Plotting these logarithmically we get the exponent of r, the distance of a planet from the sun is

-1.5 so that the density distribution of the protoplanetary disc is:

Giving a mass

With pressure gradients playing the key role in the formation of solar system, less attention is

payed to the magnetic field of the Sun. However, in the older literature, one of the pioneer’s of

this aspect found something very interesting concerning it. He was Alfven (1942). At the time

people were suggesting instead of the solar system forming from a rotating nebula, rather the

sun came into existence not at the same time at the center of the disc, but rather passed through

clouds and captured material after already existing. He figured for the captured material its

inward component v, and density , at a distance r from the sun, had to conserve mass, which

required:

He figured as the velocities of the atoms got closer to the sun, were moving then faster, collisions

would increase, and so temperature would go up, ionizing the atoms and therefore ionized, the

magnetic field becomes important. He considered for simplicity the solar magnetic field was

generated by a dipole moment , a vector quantity, and that a particle moving in the plane of

that vector with mass m and charge q, would have all of both the gravitational and magnetic

forces in that plane, so the problem becomes two-dimensional and required only the and , of

polar coordinates. The differential equations of its motion would be:

σ (r) = σ

−3/2

= 3300

M =

∫

2π

∫

σ (r)r drd θ

= 4π r

ρv

of 49 71

Equation 13.1

And,

Equation 13.2

We can integrate equation 13.2 with the boundary condition that the angular momentum of the

particle is zero at large distances from the sun to get:

Equation 13.3

And substitute it into equation equation 13.1 for to get

Equation 13.4

Which we can write

Equation 13.5

We then integrate this with respect to r with the boundary condition that at large r and get

Equation 13.6

He then notices there is another value for which . It is

Equation 13.7

This is interesting because it means the particle can never approach the Sun closer than this

value, and it depends only on the value q/m, the charge to mass ratio of a particle. He took this

as hydrogen because ionized it is a proton, for which q/m is well defined. He estimated what the

magnetic field of the Sun could have been in this earlier stage of its life, and adjusted for the fact

that hydrogen doesn’t ionize until it reaches a velocity of 5E4 m/s and found that was the

region occupied by the major planets which are Jupiter and Saturn mostly made of hydrogen

and helium.

Certainly today we don’t see the planets as having formed from material gathered by the Sun in

its journey, but rather think the Sun and planets formed at the same time from a cloud that

collapsed into a rotating flat disc. And indeed, there may be stars in the galaxy that pass through

clouds and gather material, and indeed Alfven’s equations would hold preventing ionizing

clouds from falling into their star. But we can also apply his equation to our Sun today, for which

we know a great deal about its magnetic field, which also happens to be an important thing to

study and for which we have satellites in the Lagrange points, where the Earth’s gravity cancels

··

r = −

⊙

qμ

+ mr

d(r

θ )

= −

qμ

θ = −

∫

qμ dr

qμ

··

r = −

⊙

= −

⊙

r = 0

2GM

⊙

−

r = 0

(

2GM

⊙

)

1/3

of 50 71

with that of the Sun, where the orbits are very stable, so we can understand the solar magnetic

field. It is a complex field, that interacts with the Earth’s magnetosphere, and we need to predict

solar maximums, so we have warning as to whether there will be a magnetic storm that will

knock out our electrical grid and internet, ahead of time.

During solar minimum the solar magnetic field has closed lines, that flow out one pole and into

the other. The dipole field of the sun is about 50 Gauss. There are 10,000 Gauss in a Tesla, so

that is 5E-3 Tesla. That is the magnetic field strength where the field goes into the poles. The

total magnetic field of the Sun at the Earth, is all the components taken together, which are

, , and . The important component is , because it runs north-south, so it is

perpendicular to the ecliptic, the path traced out by the sun due to the earth’s orbit. It is the

component that interacts with the Earth magnetosphere, and when it points southward, it will

connect with the Earth’s magnetosphere which points northward so the solar poles flow into the

Earth poles and the Earth field then gets disrupted allowing particles from the solar wind to rain

down along Earth magnetic field lines causing the Aurora. The solar magnetic field doesn’t

always stay around the Sun itself, but the solar wind carries it through the solar system until it

collides with the interstellar medium reaching the heliopause. Thus the Sun creates the

Interplanetary Magnetic Field (IMF) which has a spiral shape because the Sun rotates once

about every 25 days. But the upshot is that at Earth we have

Moderate Magnetic Field: 10 nT

Strong Magnetic Field: 20 nT

Very Strong Magnetic Field: 30 nT

For our purposes we want to return to equation:

of 51 71

And ask just what is , because in the time that Alfven was working we worked with magnetic

fields differently, aside from his equation uses a trick, which we still use today, and that is to

consider the magnetic field a dipole. To consider it like this is to say there are two monopoles

opposite in polarity. According to Maxwell’s equation we cannot have magnetic monopoles,

though they are predicted by some modern theories, they have never been found. The trick is in

that by treating the North magnetic pole and South magnetic pole as separate magnetic charges

is to treat them like we do electric charges, the charge of a proton and the charge of an electron,

which can be convenient for making computations, but don’t exist that we know of. So we will

solve equation 56 for , and see what its units are so we can understand what it represents and

we will let m be the mass of a proton and q the charge of a proton. We get:

Equation 13.8

We can write these units as, by taking Coulombs (C) equal to

Equation 13.9

This is units of force per current density, which makes sense because a flowing current creates a

force. We can also write it:

Equation 13.10

Which is energy per magnetic field strength in that the SI units of magnetic field strength is

amps per meter. This tells us:

Equation 13.11

Thus we will use the energy as ionization of hydrogen, the energy to remove its electron and

make it a proton:

We have

, , ,

(

2GM

⊙

)

1/3

⊙

= μ =

⋅ kg

C ⋅ s

a m p ⋅ secon ds

⋅

a m ps

⋅

a m ps

μ =

En erg y

Magnet icField Strength

H ⟶ H

+ e

−

= 1proton = 2.18E − 18J

= 1.6E − 19C

= 1.67E − 27kg

G = 6.67408E − 11N

⊙

= 1.989E 30kg

of 52 71

We want to look at the Sun as having a current flowing around its equator in a loop with its

radius

We find for the dipole field of the Sun at 50 Gauss=5E-3T, which is about 100 times stronger

than the Earth magnetic field, that this is a current I=5.5362E12 amperes driving the solar

magnetic dipole. This gives us that since the Earth orbit (1AU=1.495979E11m):

Equation 13.12

Equation 13.13

Or,…

Equation 13.14

The radius of a proton is 0.833E-15m. We have that r is:

Equation 13.15

⊙

= 6.957E8m

1AU

5.5362E12a mps

1.496E11m

= 37.0 a mps /m

μ =

Ioni zat ion En erg y

Magnet icField Strength

2.18E − 18J

37.0A mps /m

= 5.892E − 20

J ⋅ m

r =

(

2GM

⊙

)

1/3

= 4.9324E − 15m

= 5.92 ≈ 6Pr oton Ra dii = carbon

of 53 71

Our six-fold symmetry unfolding. This is again the carbon the core element of life. We see it

provided for by the Sun’s magnetic field.

of 54 71

Let us write the computation as one equation, and verify it. We have

Where

=Ionization energy of Hydrogen

=3.47E-39 (correct)

Equation 13.16

(0.000034574)(3.47E-39)=1.19972E-43

1.19972E-43^(1/3)

=4.932E-15

So as you can see equation 7.15 is correct. It says that carbon, the basis of life is in the ratio of

the solar magnetic field and the solar gravitational field.!

r =

(

2GM

⊙

)

1/3

= 4.9324E − 15m

⊙

1AU

5.5362E12a mps

1.496E11m

= 37.0a m ps /m

1AU = r

μ =

Ioni zat ion En erg y

Magnet icField Strength

2.18E − 18J

37.0A mps /m

= 5.892E − 20

J ⋅ m

⊙

1AU

⊙

= (5.982E − 20)

= 3.47E − 39

(IE

)

⊙

(2.18E − 18)

(12.56637E − 7)

(1.496E11)

4(5E − 3)

(6.957E8)

(

2GM

⊙

⋅

(IE

)

⊙

)

1/3

(1.6E − 19C )

2(6.67408E − 11N

)(1.989E 30kg)(1.67E − 27)

⋅

(2.18E − 18J )

(12.56637E − 7T ⋅ m /A)

(1.496E11m)

4(5E − 3T )

(6.957E8m)

1/3

of 55 71

Thus we have a theoretical value for the radius of the proton:

Equation 13.17

And a theoretical value for its charge

Equation 13.18

Equation 13.19

And we have the radius of a proton in terms of the solar magnetic field at Earth

Equation 13.20

All of this based on the idea that the basis of their structure is in six-fold unfolding. Given our

constant k

Equation 13.21

And making the approximation we can with these equations eliminate in equation

5.21 using equation 5.19 in which we can eliminate k with equation 2.22.

Equation 13.22

This has an accuracy of close to 88% because in equation 7.19 the charge of six protons is

predicted by the theory to be 5.72 protons, the rest of the equations are much more accurate, but

we seek to rectify that. We write this equation so we can have an equation that defines the solar

magnetic field by solving for .

To address the accuracy of the equation for the charge of a proton, equation 7.19, we ask what is

the culprit. We suggest it is . So we solve the equation for that to see by how much it is

off. It is:

Equation 13.23

We see it should be 7.79573E-11 and is actually 7.885E-11. The value we are using is 98.86785%

accurate, but we want to do better.

2π

⋅

𝔼

(

⋅

)

h 4πr

= q

= ⋅

9.1435E − 19C

1.602E − 19C

= 5.71proton s ≈ 6pr oton s

(

2GM

⊙

⋅

(IE

)

⊙

)

1/3

k =

8π

(1 + α) ⋅ N

𝔼

1 + α = 1

31850496r

⋅

𝔼)

h 4π

(IE

)

⊙

(α

)/36

= q

⋅

⋅ k

h 4πr

of 56 71

We now eliminate on the left in equation 7.22 with equation 7.17 to find the B field of the Sun

as described by the Earth orbit as the ground state:

Equation 13.24

Where N=31850496 is a perfect integer, that is has no values after the decimal. Equation 8 gives

a magnetic field strength of 4.73E-3 Teslas. Concerning equation 7.24

We want it to be more accurate. To do that we have to substitute for , the fraction

which is 39/5 and we have to formulate a theory for why this would be, though the discrepancy

could be in as its experimental value has the largest errors. Other possibilities are the

is right and that as well this is due to the radius of a proton having large errors, or even that it is

supposed to be the factor 8, which would be good to consider because 8-fold symmetry is very

dynamic, in particular in its role with beryllium 8 being a precursor to carbon in nuclear

synthesis by stars.

Because we are looking at 6 protons we had , and because of six-fold symmetry we had

and because the charge we had because of the 1/3 root on the right. So taking

and leaving the that came from substituting for k, and the as

well, and leaving the factors 2 and 4 because they describe the physical dynamics of the equation

we have:

Equation 13.25

Where

Or we can factor out all the numbers on the right the 2 and 4 in equation 7.25 and write

Equation 13.26

In which case N=31850496/4=7962624. We see the solar magnetic field is determined by the

radius of a hydrogen atom, its ionization energy, and the solar gravitational field, with the earth

orbit as the ground state.!

⊙

⋅

(IE

)

⊙

⋅

h 4π

= q

⋅

⋅ k

h 4πr

(α

)/36

(α

)/36

q = 6q

N = 6

= 279936

8 = 2

⊙

⋅

(IE

)

2GM

⊙

⋅

h 4π

N = 6

= 279936

⊙

⋅

(IE

)

⊙

⋅

h π

of 57 71

Protons and neutrons pack in atoms in a way that can be seen from their electron clouds. Thus

since we can say the electron cloud of helium is spherical the nucleus packing is spherical. But

since the protons and and neutrons in the nucleus are moving (But only as far as their nuclear

wall) the packing is ordered but liquid.

I believe the reason for my finds of six-fold symmetry in both the radius of a proton and it its

charge can be explained by Buckminster Fuller’s vector equilibrium. It is the most

transformable straight-line geometry if you attach sticks with flexible corners, what Fuller calls a

flex-corner. See the following illustration.

Which means it zero-frequency for omni directional closest packing of spheres is 12 spheres,

which is carbon (six protons and six neutrons) which is the basis of life as we know it. Its

diameter is then six proton radii which we found provided for by the solar magnetic field. The

frequency is the number on any symmetrically concentric shell or layer and is given by

The more protons an element has the more neutrons it needs so the strong nuclear force in the

nucleus can overcome the mutual repulsion of protons, thus holding the atoms together.

10F

+ 2 = 10(1)

+ 2 = 12 = carbon

of 58 71

Thus, to go over that of the solar magnetic field again, proposed by Hannes Alfven (1942), at the

time there was no known mechanism for it, but he suggested the relative velocity between a

neutral gas and a plasma has a critical velocity at which the gas starts to ionize and that the

atoms or molecules will not exceed this velocity until the gas becomes almost fully ionized. The

additional energy put into the system goes into ionizing the gas instead of the velocity of the

atoms, and is roughly independent of pressure and magnetic field. Critical ionization velocity

has been recognized in the laboratory for some time. It is given by equating the kinetic energy of

the atoms to the ionization potential:

Alfven found:

Gas cloud enters Solar System

A neutral atom falls into Sun due to gravity

Motion is random, collisions happen

Temperature rises

At a distance from the Sun gas will ionize

m=atomic weight

He found for a gas cloud with average voltage 12 volts. An average atomic weight of 7, the is at

Jupiter.

That is atoms fall in towards the Sun and ionize at which point the solar magnetic field pushes

them out to Jupiter orbit where a halo forms from which planets can form.

= eV

ion

= G

ion

= G

ion

= 13.5E10

ion

= volts

of 59 71

Thus if we are to suggest, as we have, that solar magnetic field provides for life, whose skeletons

are the hydrocarbons, that this has something to due with the solar wind which is plasma and

we can consider it a plasma that can move in neutral gas thus being subject to Alfven’s critical

velocity. The plasma is mostly electrons, protons, alpha particles kinetic energies between 0.5

and 10 keV. There are trace amounts of heavy ions of C, N, O, Ne Mg, Is, S, and Fe. We note the

C, N, O are the most abundant elements in life chemistry. The solar wind can reach velocities of

25o000-750000 m/s. Remember the critical velocity for hydrogen is 50,900 m/s.

of 60 71

14.0 Eightfold Symmetry Since and is dimensionless, we can write in the 6 of sixfold

symmetry for it:

We see that

Equation 14.1.

Where hydrogen and carbon are the central structural elements of life composing the

hydrocarbons which are the skeletons of biological chemistry.

Thus

Is a quantity that maps things to things. We find since the orbital velocity of the moon

, we have

We have the equation

Which honed is

k v

= 6

⋅

h 4π

K E

moon

K E

earth

(Ear th Da y) k v

⋅

h 4π

= 6

K E

moon

K E

earth

(Ear th Da y)

4πh

↦ 6secon d s

⋅

h 4π

= 6proton s = Carbon

⋅

h 4π

= 1proton = Hydr ogen

(

)(

4πh

)

(

4.98E11

)(

1.21E − 11kg

)

= 6secon d s

4πh

= 1.21E − 11kg

= 1,022m /s

(1.21E − 11kg ⋅

)(1022m /s) = 1.23662E − 8kg

ℏ

1/2

of 61 71

which gives gives

Equation 14.2.

Thus we have

Six-fold which is carbon (C) has map the radius to mass of a proton to the kinetic

energy of the Moon to the kinetic energy of the Earth by a factor of the Earth period of rotation,

which is 1 second and that the intermediate mass , is 8 times the factor that does that

mapping.

Equation 14.3.

Where 1 second is

Giving the radius of a proton as

From

Where is six seconds. What is the comparison of six-fold symmetry to eight-fold symmetry?

Six-fold is in the 18 groups of the periodic table that organizes elements according to their

properties. Six-fold is dynamic because 2 and 3 are the smallest prime numbers so they are the

smallest factors down to which any number can be reduced. We have

, , ,

= (67.9943)(1.01328) = 68.897kg ≈ 69kg ≈ 70kg

68.897kg

1.23662E − 8kg

= 8.0865585 ≈ 8

(

4π

)

= 8

⋅

h 4π

K E

moon

K E

earth

(Ear th Da y) = 1secon d

4πh

= 8

(

4π

)

3 2

⋅

Gπ

⋅

𝔼

= 1secon d

2π

⋅

𝔼

⋅

h 4π

= 1proton

2 ⋅ 3 = 6

= 9

2 ⋅ 9 = 18

3 ⋅ 6 = 18

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That is dynamic. That is how you play music in 6/8 that resolves in . It is in the

Flamenco of Spain to play 3 groups of six to make 18 like in the periodic table of the elements,

then strum another group of chords six beats long usually with rhythm 2+2+2, to make 24 in

order to come around to cycles of 12. This form is called Bulerias, and is considered the most

dynamic in the genre. However, just as popular is eight, the tango flamencos, and the rumba,

which is 4+4+4+4=16. Other forms are fandangos (four groups of 3 making 12) and Sevillanas,

the same thing. Middle Eastern music has much along these lines and so does Hindu Indian

music which are North Indian Classical music (Mostly tin tal which is four groups of four to

make 16) and Ghuzals, romantic music played in 6.

12 = 2 ⋅ 6

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Appendix 1 Kinetic Energies Moon and Earth

We suggested the second was a natural unit, and that, if it was, should be in the orbital

mechanics of the earth because the second comes from the calendar, which is based on the

orbital period of the year (1 year, 365.25 days) and the orbital period of the moon, and the

rotation of the Earth. We found that it was as the following:

Let’s show that…

To find the translational kinetic energy of the moon:

Distance from earth: 3.85E8m

Orbital period:

T=27.32 days=2.36E6seconds

v=1.025E3m/s~1000m/s

Mass: 7.34767E22kg

Use

E=3.67E28 Joules

To find the translational kinetic energy of the earth:

Distance from Sun: 1AU=1.496E11m

Orbital period: 1 year=3.1558E7 seconds

v=2.9785E4m/s, Earth mass: 5.9722E24kg

E=2.649E33 Joules

Earth day=(24)(60)(60)=86,400 seconds.

Using the Moon’s orbital velocity at aphelion, and Earth’s orbital velocity at perihelion we have:

Using the Moon’s orbital velocity at aphelion, and Earth’s orbital velocity at perihelion.

K E

moon

K E

earth

(Ear th Da y) ≈ 1secon d

K . E . Moon = 3.67E 28J

K . E . Ear th = 2.649E 33j

2π (3.85E 8m) = 2.419E 9m

E =

2π (1.496E11m) = 9.399E11m

K E

moon

(7.347673E 22kg)(966m /s)

= 3.428E 28J

K E

earth

(5.972E 24kg)(30,290m /s)

= 2.7396E 33J

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Appendix 2 Van Der Waals Radius

Johannes Diderik van der Waals (1873) described more than just Ideal gases, which are gases

that behave according to kinetic-molecular theory, he described real gases which don’t. His

equation then, The Van der Waals equation, is a modification of the Ideal Gas Law which is:

Which is quite obvious. If you increase the temperature T, then the volume of the gas is going to

increase, and if it doesn’t then the pressure will, which is inversely proportional to volume.

However for a Real Gas, he assumed the particles are hard spheres, cannot be compressed

beyond a limit, and at close proximity to one another they interact and have a volume around

them that excludes one another, that is they have walls. He said

That is the volume of the real gas ( ) is equal to the volume of the ideal gas ( ) minus a

correction factor b. The volume of the particles is the number of particles ( ) times the volume

of one particle:

Thus there exists a sphere of radius 2r formed by two particles in contact where no other

particles can enter. It gives the correction factor

PV = n RT

= V

− b

π r

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And the volume correction for n particles is

This is the volume correction to the Ideal Gas Law. The pressure correction says real gases

exhibit less pressure because their particles interact which is a net pulling by the bulk of

particles away from the container walls.

The reduction in pressure is proportional to by a factor a. We have for reduction of pressure

that

We substitute this into the Ideal Gas Law:

This can be written as a cubic

Which allows one to compute the critical conditions of liquefaction and to derive an expression

of the principle corresponding states. In the cubic form we have as the solution three volumes

which can be used for computing the volume at and below critical temperatures.

Thus the Van Der Waals radius is estimated

For hydrogen experimentally. Therefore with

We have described the derivation of radius of a hydrogen atom from the Van Der Waals

equations that we use to get

b = (4)

π r

nb = 4n ×

π r

= P

+ a

(

P + a

)

(

V − nb

)

= n RT

−

(

b +

)

V −

a b

= 0

(

π +

)

(3φ − 1) = 8τ : π =

, φ =

, τ =

π r

4π

⋅

b = 26.61

m ol

= 6.02E 23

= 1.0967E − 8cm = 1.0967E − 10m

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Where the Van Der Waals equations are

π α

= 1secon d

(

P + a

)

(

V − nb

)

= n RT

−

(

b +

)

V −

a b

= 0

π r

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Appendix 3 Chandrasekhar Limit

The pressure of the outer shell of star balances with the outward pressure in the core of

the star (thermal pressure). Pressure is force per unit surface area thus…

is the mass of the core pulling in the mass of the shell and is the radius of the

core. The surface area of the star is that of a sphere, . We have

The thermal pressure countering the gravity is given by the ideal gas law PV=nRT (pressure

times volume of a gas such as hydrogen , which is all protons , is proportional to temperature.

The number of protons in the core is . We have

Where is the Boltzmann constant ( ). Since we must have

if the star is not to implode or explode

And we have the estimate for the temperature of the core of a star.

Fusion would not occur at the low temperature of a star like the Sun in that there would not be

enough energy for collisions, unless the potential Coulomb barrier can be overcome by quantum

mechanical tunneling. The collisions are given by the kinetic energy of the particles

. We have

The velocity v yields the minimum distance between protons as the De Broglie wavelength

gravit y

P =

F = m a = PA

P =

m a

m a = G

shell

core

shell

core

A = 4π r

core

gravit y

= G

shell

core

4π r

core

ther mal

core

⋅

π r

core

1.380649E − 23J ⋅ K

−1

gravit y

= P

ther mal

core

shell

core

4πϵ

min

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Since the velocity is the root mean square velocity of the protons…

We have the temperature of the star is

This is another estimate. Since the mass of a star is its volume times its density

But for a star density varies with radius

If we take the derivative of both sides of the equation we have one of the equations of stellar

structure:

The so-called conservation of mass equation. The force on the shell of the star is given by the

mass of the shell

Again for there to be balance gravitation pressure equals thermal pressure:

Another equation of the equations of stellar structure. The so-called equation of hydrostatic

equilibrium. This can be written

If the star is an ideal gas the density of the star varies as where for a

monatomic gas and then

λ =

rms

min

(

4πϵ

)

3π

m =

π r

4π

∫

ρ(r)dr

dm(r)

= 4π r

ρ(r)

= − G

m(r)4π r

ρ(r)

dρ(r)

= − G

m(r)ρ(r )

(

ρ(r)

dP(r)

)

= − 4π Gr

ρr

= consta nt

γ = 5/3

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In stellar dynamics we write

So that

The abundance of hydrogen and helium in the universe are approximately 75% and 24%,

respectively. Thus for every 4He2+ there are 12H+ and 2+12 free electrons. We have

Ionized hydrogen and helium have and for the Sun because of high metal

content. Finally stars can be approximated as blackbody radiators (purely radiate) and as such

pressure is given in terms of temperature (Temperature is proportional to radiation energy):

There are three kinds of pressures that can be generated by a star: gas pressure, radiation

pressure, or degeneracy pressure.

A type of star that is stable, that is prevented from collapse by degeneracy pressure, is a so-called

white dwarf star. They are the remnant of giant stars that have depleted the their fusion fuel and

thereby collapsed under gravity but are kept from collapsing into black holes by thermal

pressure due to motion of the particles alone. Interestingly, they still shine almost as bright as a

star on the main sequence even though they are not doing fusion. It was the Indian physicist

Chandrasekhar who found the limit in mass for which a white dwarf will not have its gravity

overcome the degeneracy pressure and collapse. The non-relativistic equation is:

There are many resources available that derive this and you can find it in any textbook on

astrophysics in the chapters dealing with stellar physics, and I will leave the treatment of the

derivation to those works.

P ∝

∝ ρ

5/3

μm

gas

μm

4 + 12

1 + 12 + 14

= 0.59

μ = 0.59

μ = 0.62

rad

⋅

M ≤ 0.77

= 1.41 ⊙

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The Author